Engineering 9867 Advanced Computing Concepts

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1 Engineering 9867 Advanced Computing Concepts Assignment #2 Sample solutions Due: Tuesday, April 2 at 9. [ points] Consider the following implementation of the palindrome checking problem (question 4 on assignment ): bool ispalindrome(const string& s) { bool result = true; int size = s.size(); int i = ; } while (result && i < size/2) { result = (s[i] == s[size--i]); i++; } return result; a) [5 points] Give an expression for the worst case time (note not complexity) of this algorithm. The worst case time occurs when s is a palindrome, in which case the loop repeats size/2 times, so WT ispalindrome (size) = k + k size/2 b) [5 points] Assuming that the strings contain only of the letters a and/or b, and that all strings are equally likely, what is the average case time for this algorithm? Let N be the length of the string. For strings containing only a or b, the probability any particular comparison will find the letters unequal, and thus cause the algorithm to stop is P(s[i] s[n i]) =. The probability that it will stop at exactly { 2 the i th, for i < N itteration P(stop at i) = 2 i 2 i j= P(stop at j) =, for i = N and 2 i 2 thus the expected number of itterations is E(N) = N 2 j= j P(stop at j) so AT ispalindrome (N) = k 2 = N N 2 j= 2 N 2 j 2 j

2 2. [ points] Consider the following algorithm that satisfies the specification as follows: Pre: m n Post: result n m ( i, i < m P[i] = S[result + i]) result > n m ( j, j n m ( i, i < m P[i] = S[j + i])) result = matched = false while (result < n m matched) do result = result + i = matched = true while (i < m matched) do matched = matched (P[i] == S[result + i]) i = i + end while end while if ( matched) then result = result + end if a) [9 points] Give an expression for the exact worst case (i.e., Θ) complexity for this algorithm. Show your workings. In the worst case the outer loop will execute n m times, and each time the inner loop will execute m times (this would not actually happen, but it doesn t make any difference for the complexity), so WT match (n, m) Θ(n m). b) [ point] What does this tell us about the complexity of the problem solved by this algorithm? This tells us that the problem of pattern matching is O(n m) (i.e., the problem complexity is no worse than linear in the length of the string to search). Note that this does not tell us that the complexity is no better than this, in fact there are sub-linear algorithms for this problem Assignment 2 2 Revised: April 6, 22

3 3. [5 points] Give a (deterministic) finite state automata on the input language Σ = {, } accepting each of the following languages: a) [5 points] The set of all strings ending in. b) [5 points] The set of all strings with three consecutive s., 9867 Assignment 2 3 Revised: April 6, 22

4 c) [5 points] The set of all strings such that every block of five consecutive symbols contains at least two s. _ 9867 Assignment 2 4 Revised: April 6, 22

5 4. [5 points] Let G be an undirected graph consisting of a set of nodes, N, and a set of edges E N N. A set of nodes N N is called a vertex cover for G, if for every edge in E, at least one of its end-points is in N. The vertex cover problem is, given a graph, G, and a positive integer, K, determine whether there is a vertex cover for G with at most K nodes. Show that the vertex cover problem is NP-complete Assignment 2 5 Revised: April 6, 22

6 First we must show that VCP is in NP. To do this we give the non-deterministic polynomial time algorithm for VCP, as follows: Assume that N = {n, n,... n M }, for some M > K, and that E = {e, e,... e Q }. for i = to M do if (magiccoin() N < K) then Add n i to N end if end for for i = to Q do if Neither of the endpoints of e i N then output No stop end if end for output Yes Clearly this algorithm is polynomial time, so VCP is in NP. Now we must show that another problem that is known to be NP-complete can be reduced to VCP. I ll use 3SAT. For an instance of 3SAT, assume we have the variables v, v 2,... v n, and the 3CNF formula (a, a,2 a,3 ) (a 2, a 2,2 a 2,3 )... (a m, a m,2 a m,3 ) where each a i,j is v k or v k. We construct { an instance of VCP as follows: The set} of nodes in G is v, v N = 2,... v n, v, v 2,... v n, and the set of edges is a,, a,2, a,3, a 2,, a 2,2, a 2,3,... a m,, a m,2, a m,3 (v, v ), (v 2, v 2 ),... (v n, v n ), (a E =,, a,2 ), (a,2, a,3 ), (a,, a,3 ), {(a (a 2,, a 2,2 ), (a 2,2, a 2,3 ), (a 2,, a 2,3 ),... i,j, v k ) a i,j is v k } {(a i,j, v k ) a i,j is v k } (a m,, a m,2 ), (a m,2, a m,3 ), (a m,, a m,3 ) That is, G consists of an edge for each variable and its negation, a triangle for each clause and an edge connecting each term in the clause to its value. Choose K = n + 2m the number of variables plus twice the number of clauses. The formula is satisfiable iff there is a vertex cover of G containing at most K nodes. This is illustrated in the following figures. A!A B!B C!C A!A B!B C!C (A+B+C)&(!A+!B+!C) (A+B+C)&(!A+B+C) 9867 Assignment 2 6 Revised: April 6, 22